The profit for a product can be described by the function \( \mathrm{P}(\mathrm{x})=212 \mathrm{x}-5000-\mathrm{x}^{2} \) dollars, where x is the number of units produced and sold. To maximize profit, how many units must be produced and sold? What is the maximum possible profit? To maximize profit, 106 units must be produced and sold. (Simplify your answer.) The maximum possible profit is \( \$ \square \) (Simplify your answer.)

To determine the maximum possible profit, substitute \( x = 106 \) into the profit function \( \mathrm{P}(\mathrm{x}) = 212\mathrm{x} - 5000 - \mathrm{x}^2 \): \[ \mathrm{P}(106) = 212 \times 106 - 5000 - 106^2 \] Calculate each term step-by-step: 1. **Calculate \( 212 \times 106 \):** \[ 212 \times 100 = 21,200 \] \[ 212 \times 6 = 1,272 \] \[ 21,200 + 1,272 = 22,472 \] 2. **Calculate \( 106^2 \):** \[ 106 \times 106 = 11,236 \] 3. **Substitute the values back into the profit function:** \[ \mathrm{P}(106) = 22,472 - 5,000 - 11,236 \] \[ \mathrm{P}(106) = 22,472 - 16,236 = 6,236 \] **Therefore:** - **Number of units to produce and sell for maximum profit:** 106 units - **Maximum possible profit:** \$6,236 \(\boxed{\$6,\!236}\)